A hair salon claims that 90% of their customers are satisfied.However,Toni who recently started working there found that of 120 customers last week,only 100 were satisfied.If it were true that on average 90% of customers are satisfied,what is the chance that among 120 customers 100 or fewer would be satisfied?
A)0.2297
B)0.9925
C)0.7703
D)0.0075
E)0.0003

Question

Quizplus · Accepted Answer

We can model the number of satisfied customers among 120 as a binomial distribution with n=120 and p=0.9, assuming that each customer's satisfaction is independent of others. Then, we want to find P(X ≤ 100), where X is the number of satisfied customers. We can use the normal approximation to the binomial distribution by using the continuity correction term (adding/subtracting 0.5) since n is large enough and p is not too extreme. Then, we have:

µ = np = 120*0.9 = 108
σ = sqrt(np(1-p)) = sqrt(120*0.9*0.1) = 3.06

P(X ≤ 100) ≈ P(Z ≤ (100+0.5-108)/3.06) 
              = P(Z ≤ -2.45) 
              ≈ 0.0075  (using a standard normal table or calculator)

Therefore, the chance that among 120 customers 100 or fewer would be satisfied is about 0.0075, which corresponds to answer choice D.

A Hair Salon Claims That 90% of Their Customers Are